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Projectile motion 
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#1
Sep903, 12:00 AM

P: n/a

I need help putting this problem into a workable equation:
A projectile is launched from ground level at an angle of 12 degrees above the horizontal. It returns to ground level. To what value should the launch angle be adjusted, w/o changing the launch speed, so that the range doubles? So far I tried this: I broke the problem into its components: x & y. FOR X: {x=Vo*COS(12)t Vox=Vo*Cos(12) a=0 t=t} FOR Y: {y=? Voy=Vo*SIN(12) a=9.8 m/s^2 t=t} Then I used t=x/Vo*COS(12), then substituted that for T in the Y parts, so that: y= tan(12)x  [(4.9 x^2)/(Vo^2*cos(12)^2)] but then I got: x=tan(12)y, which doesn't help me at all. I'm guessing I'm approaching this problem in a totally WRONG way! 


#2
Sep903, 05:07 AM

Emeritus
P: 1,919

so you start out with a vector v> and an angle theta1.
x = v> * cos(theta1) * t y = v> * sin(theta1) * t  (1/2) * g * t^2 when y = 0, the projectile will be on the ground 0 = v> * sine(theta1) * t  (1/2) * g * t^2 t = 0 , 2 *v> * sin(theta1) / g when t = 0 is at the start, so the other solution is the time it hits the ground x = v> * cos(theta1) * 2 *v> * sin(theta1) / g pluging in t into the x equation and doing some math you get: x = v>^2 * sin( 2 * theta1) / g If you want the range to be double then it is 2 times the x equation with the new angle. x2 = 2 * v>^2 * sin( 2 * theta2) / g equate x and x2 to find theta2. You know theta1 = 12 degrees x = x2 , then theta2 = 5.87 degrees And there is your answer 


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